Home > Calculus > Derivative Rules for y=cos(x) and y=tan(x)

How do you find the derivative of #w=1/sinz#?

Answer 1

#(dw)/(dz)=-cos(z)/(sin^2(z))#

For the general case, the derivative quotient rule tell us: #color(white)("XXX")(d (f_x))/(d (g_x)) =((df_x)/(dx) * g_x - (dg_x)/(dx) * f_x)/(g_x^2)#

Taking #f(z) = 1# and #g(z)=sin(z)# (so #w_z=(f_z)/(g_z)#) and remembering that #color(white)("XXX")(d sin(z))/(dz)=cos(z)# we have #color(white)("XXX")(dw_z)/(dz)=(0 * sin(z) - cos(z) * 1)/(sin^2(z))#

#color(white)("XXXXX")=(-cos(z))/(sin^2(z))#

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Answer 2

Ezekiel Alexander

#(dw)/dz = -csc(z) cot(z) = (-cosz)/(sin^2z)#

The answer below is also valid, but here is a shortcut if you can remember the identity:

#w = 1/sinz = cscz#

#therefore (dw)/dz = d/dz csc z = -csc(z) cot(z)#

So the answer can be written as #-csc(z) cot(z)# or #(-cosz)/(sin^2z)# since they are equivalent forms.

Final Answer

This is a common basic identity which can be quickly memorized along with d/dx of sine, cosine, tangent, etc. in order to save time in the future:

#d/dx csc(x) = -csc(x) cot(x)#

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Answer 3

Emma Aldridge

To find the derivative of ( w = \frac{1}{\sin(z)} ), you can use the quotient rule. The derivative can be expressed as:

[ \frac{dw}{dz} = -\frac{\cos(z)}{\sin^2(z)} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.